# Class 12 Physics Sample Paper

We have provided Class 12 Physics Sample Paper as per the latest CBSE examination pattern for the current academic year. The following CBSE Sample Papers for Class 12 Physics has been prepared based on the guess papers issued recently. Students will be able to practice these papers and get good marks in upcoming Physics exams for Class 12.

## Class 12 Accountancy Sample Paper Term 1 With Solutions Set A

SECTION-A

1. Mention the important considerations required while fabricating a p-n junction diode to be used as a Light Emitting Diode (LED). What should be the order of band gap of an LED if it is required to emit light in the visible range?
Answer : (i) The semiconductors used for fabrication of visible LEDs must have a band gap from 1.8 eV to 3 eV.
(ii) Both p-type and n-type semiconductor are heavily doped.
(iii) It should be encapsulated with a transparent material.
For emitting light in visible range the band gap should be from 1.8 eV to 3 eV.

2. A monochromatic source, emitting light of wavelength, 600 nm, has a power output of 66W. Calculate the number of photons emitted by this source in 2 minutes. Use h = 6.6 × 10–34 Js.
Answer : Here, l = 600 nm = 6 × 10–7 m; P = 66 W,
N = ?, T = 2 × 60s

Energy of each photon,

E = hc/λ = (6.6 x 10-24) x (3×108) / 6×10-7 = 3.3 × 10–19 J

Energy emitted by source in time t,

ES = Pt = 66 × 2 × 60 J
\ No. of photons emitted by the source in 2 minutes
N = Eg/E = Pt/E = 66 x 2 x 60 /3.3 x 10-19 = 2.4 x1022 photons.

3. Distinguish between intrinsic and extrinsic semiconductors on the basis of energy band diagram.

SECTION-B

4. Radiations of frequency v1 and v2 are made to fall in turn, on a photosensitive surface. The stopping potentials required for stopping the most energetic photoelectrons in the two cases are V1 and V2 respectively. Obtain a formula for determining the threshold frequency in terms of these parameters.
Answer : If v0 is the threshold frequency, then from photoelectric equation, we have

eV1 = hv1 – Φ0 and eV2 = hv2 – Φ0
e(V2 – V1) = h(v2 – v1) or h = e(V2 – V1)/(V2 – V1)
Now eV1 = hv1 f0 = hv1– hv0
Or v0 = v1 – eV1/h = v1– eV1 [v2 – v1 /e(V2 – V1)]
= v1 – V1(v2 – v1) /(V2 – V1) = v1V1 – v1V1 – v2V1 + v1V1 /(V2 – V1) = ( v1V2 – v12V1 )/(V2 – V1)

5. Two material bars A and B of equal area of cross-section, are connected in series to a DC supply. A is made of usual resistance wire and B of an n-type semiconductor.
(a) In which bar is drift speed of free electrons greater?
(b) If the same constant current continues to flow for a long time, how will the voltage drop across A and B be affected?
Answer : (a) Drift speed in B (n-type semiconductor) is higher
Reason: I = neAvd is same for both
n is much lower in semiconductors.
(b) Voltage drop across A will increase as the resistance of A increases with increase in temperature.
Voltage drop across B will decrease as resistance of B will decrease with increase in temperature.

6. Calculate and compare the energy released by
(a) fusion of 1 kg of hydrogen deep within the sun and
(b) the fission of 1 kg of 235U in a fission reactor.
Answer : (a) In sun 4 hydrogen nuclei combine to form a helium nucleus and 26 MeV of energy is released (Img 8)

No. of atom in 1g of hydrogen = 6.023 × 1023
No. of atom in 1kg of hydrogen = 6023 × 1023
Energy released by 1 kg of hydrogen = 26 x6023x1023 /4 ≃39×1026 MeV

(b) No. of atom in 235g of
235U = 6.023 × 1023
No. of atom in 1 kg of
235U = 6.023 x1023 x 1000/235
Energy released in per fission of
235U = 200 MeV
Energy released in per fission of 1 kg of 235U
6 023 x 1023 x 1000 x 200/235 ≃ 5.1 x 1026 . × M V

7. A convex lens made up of glass of refractive index. 1.5 is dipped in turn in (i) a medium of refractive index 1.65 (ii) a medium of refractive index 1.33 (a) will it behave as a converging or diverging lens in two cases (b) How will its focal length changes in two media.
Answer : (i) refractive index of medium = m1 = 1.65, m2 = 1.5,
R1 = R, R2 = – R
(Img 9)
It will behave as a diverging lens as f is –ve.
If m1 = 1.33 then 1/f = (1.5/1.33 – 1)2/R = .17/1.33. 2/R
f is + ve so converging lens.
(b) In air 1/fa = (1.5-1)2/R = 1/R or f = R

In medium of refractive index 1.65,
f = -165/15 R/2 = -5.5R
In medium of refractive index 1.33,
f = 133/17 R/2 = 7.8/2R = 3.9R
f is maximum in medium of r-I 1.65.

8. Draw a schematic diagram of a reflecting telescope (Cassegrain). Write two important advantages that a reflecting telescope has over a refracting telescope.
Answer : Objective is concave mirror of large focal length and large aperture having a hole around its pole. The light coming from a far off object after reflection from the objective is made to fall on a secondary mirror (convex)placed in between its pole and focus. The light reflected from the convex mirror goes into the eye piece, which acts as a simple microscope.
Advantage of reflecting telescope over a refracting telescope: (Img 9)
(i) Reflecting telescope is free from chromatic aberration.
(ii) Image is free from spherical aberration.
(iii) Image formed is brighter and resolving power is high.

9. How is a wavefront defined? Using Huygen’s construction draw a figure showing the propagation of a plane wave reflecting at the interface of the two media. Show that the angle of incidence is equal to angle of reflection.
Answer : Consider a plane wavefront AB reflected at the interface of the two media. When the wavefront strikes the reflected surface at point B the secondary wavelets travel back into the some medium.
During the time the disturbance from A reach the point C, the secondary wavelets from B have spread over a hemisphere of radius BD = vt. Where v in the speed of the wave and t is the time taken from the wave to reach from A to C. (Img 9)
Also AC = vt
The tangent plane CD is the new reflected wavefront.
Let i and r be the angle of incidence and angle of reflection
respectively.
∠ABC = ∠i, ∠BCD = ∠r

In DABC and DDCB, ∠BAC = ∠CDB = 90°
BC = BC common
AC = BD = vt
So DABC ≅ DDCB
So by cpct ∠ABC = ∠BCD or ∠i = ∠r

10. A monochromatic radiation of wavelength 975 Å excites the hydrogen atom from its ground state to a higher state. How many different spectral lines are possible in the resulting spectrum? Which transition corresponds to the longest wavelength amongst them?
Answer : Energy corresponding to the given wavelength:
E (in eV) = 12400/ λ(in A) = 12.71 eV
The excited state: En – E1 = 12. 71
-13.6/n2 + 13.6 = 12.71
n = 3.9 ≈ 4
Total no. of spectral lines emitted : n(n-1)/2 = 6
Longest wavelength will correspond to the transition
n = 4 to n = 3

11. Identify the following electromagnetic radiation as per the wavelength given below:
(a) 10– 4 nm (b) 10–3 m (c) 1 nm.
Write one application of each.
Application : Radio therapy or to initiate nuclear reactions.
(b) 10–3 m → microwaves
(c) 1 nm → X-rays
Application : In medical science for detection of fractures in bones.

SECTION-C

12. Radio waves are produced by the accelerated motion of charges in conducting wires. Microwaves are produced by special vacuum tubes. Infrared waves are produced by hot bodies and molecules also known as heat waves. UV rays are produced by special lamps and very hot bodies like sun. (Img 10)

Based on the above facts, answer the following questions:

(a) transverse em wave
(b) longitudinal em wave
(c) both (a) and (b)
(d) none of these

(ii) The cause of greenhouse effect is
(a) infrared rays
(b) ultraviolet rays
(c) X-rays

(iii) Biological importance of ozone layer is
(a) it stops UV rays
(c) its layer reduces greenhouse effect
(d) none of these

(iv) Ozone is found in
(a) stratosphere
(b) ionosphere
(c) mesosphere
(d) troposphere